Tangent Plane in 4-dimensional Euclidean space.

the problem is to find the linear equations of the tangent plane in $\displaystyle \mathbb{R}^4$ to the graph of F at the point $\displaystyle (\pi/2, \pi/2,0,0)$

I am confused because the image of F has two components, so how could it have a four-dimensional tangent plane? Wouldn't there be infinetly tangent planes in $\displaystyle \mathbb{R}^4$? We can't even evaluate F at the given point...

Sorry I am so hopelessly confused - any clarity or guidance much appriciated.

The graph of the function y= f(x) is a curve in $\displaystyle R^2$ with points denoted by (x, y). The graph of the function z= f(x,y) is a surface in $\displaystyle R^3$ with points denoted by (x, y, z). The graph of the vector function (u,v)= f(x,y) is in $\displaystyle R^4$ with points denoted by (x, y, u, v).

But you are right about evaluating F at the "given point": at $\displaystyle x= \pi/2$, $\displaystyle y= \pi/2$, $\displaystyle F(x,y)= F(\pi/2, \pi/2)= (sin(\pi/2-\pi/2, cos(\pi/2- \pi/2))= (sin(0), cos(0))= (0, 1)$, NOT (0, 0) as implied by $\displaystyle (\pi/2, \pi/2, 0, 0)$. Are you sure you have copied that correctly? If it were f(x,y)= (sin(x-y), cos(x+y)) then $\displaystyle f(\pi/2,\pi/2)= (0, 0)$ would be correct.

If we think of the curve y= f(x) as a "level curve" of F(x,y)= y- f(x)= 0, then the two dimensional vector $\displaystyle \nabla F= -f'(x)\vec{i}+ \vec{j}$ is normal to the curve. If we think of the surface z= f(x,y) as a "level surface" of F(x,y,z)= z- f(x,y), then the three dimensional vector $\displaystyle \nabla F= -f_x\vec{i}- f_y\vec{j}+ \vec{k}$ is normal to the level surface. Finally, if we think of the graph (u,v)= f(x,y) as a level curve of $\displaystyle F(x,y,u,v)= (u,v)- f(x,y)$, then $\displaystyle \nabla F= -f_x\vec{i}- f_y\vec{j}+ \vec{k}+ \vec{l}$ will be normal to it.